che 333 class 18 strengthening of metals.. strengthening at cold temperatures metals – basically...
TRANSCRIPT
CHE 333 Class 18
Strengthening of Metals.
Strengthening at COLD temperatures
Metals – basically all work in same way
which is to block dislocations or retard
Them.
Remember – COLD is less than 0.3 Tm in
Kelvin
Strengthening Mechanisms
To optimize properties of metals, greater strengths can be achieved by several
techniques:-
1. Cold Working.
2. Grain Size Control
3. Solution Strengthening
4. Second Phases.
5. New Phases.
These are the engineering alloys that are used for structural applications. Pure materials
are used for electronic and electrical applications or chemical applications.
Cold Working
As the number of dislocations increase they interact and block each other. The first dislocations will bethe ones nearest 450 to the applied stress where theresolved shear stress is greatest. Therefore whenanother slip system needs to be activated, the applied stress must be increased to reach the critical resolved shear stress on a new slip plane. So to increase strain, the stress must be increased. Plastic deformationresults and so work hardening occurs and the yield stress effectively increased, strengtheningthe metal.
Dislocation Interactions1
2
2
3
Grain Size Control.Grain boundaries block dislocation motionas they change the orientation of slip planeswith respect to the applied stress. As the firstslip system activated will be the one nearest 45o then all others will need more applied stressto reach the critical resolved shear stresseson planes which are not near 450 to the applied stress.In the figure, a dislocation is blocked by the grainboundary as the (111) planes in the next grainare not at 450 to the stress applied. This will leadto a “dislocation pile up” where many dislocationsget blocked on the slip plane. This “pile up” creates a stress build up in the next grain, addingto the applied stress and so initiating slip in thenext grain. The smaller the grain size, the fewerdislocations in the pile up and the higher theapplied stress to cause further slip. So the smallerthe grain size the higher the mechanical strength
(111)
45o
30o
Hall Petch Equation
Empirical Equation relating yield
strength to grain size.
y = o + kd -1/2
y = yield stress for polycrystaline
o = yield stress single crystal
k = constant
d = grain size.
The smaller the grain size the higher
the yield stress. Grain size can be controlled
by recrystallization and other techniques.
Solution Strengthening.
Add a solute to the metal, such as zinc to copper to create brass. The zinc atoms are
a different size and so affect dislocations.
Around a dislocations stress and
strain fields exist, compressive
above the slip plans and tensile
below it. A small atom can reduce
the compressive stress field
while a large atom can reduce
the tensile stress field. The applied
stress to move a dislocation will
therefore increase if the internal
stress field is decreased. The limit for
this is the Hume Rothery rules which
limit the amount of solute before
second phases form.
Compressive
Tensile
Dislocation Locking
For steels, there is an upper andlower yield point. This is due to carbon in interstitial sites “locking”the dislocations in place. The smallatoms reduce the strain energy of adislocation. It then requires moreexternal energy in the form of stressto move the dislocation.Once the dislocation is free from thethe local carbon atom, less stressis required to move it.
Dislocation FrictionSolute atoms have strainfields associated withthem. As a result, asdislocations move past solute atoms, the energyof the dislocation islowered and more stressis required to keep it in motion. This increasesthe UTS of a material but not the yield stress.
Stress
Dislocation friction raises plastic deformation curve
Strain
Second Phases.The presence of second phases will strengthena material by blocking dislocation motion, and requiring increased applied stress to produce strain.Second phases all work in the same mannerby blocking dislocation, their effectivenessdepends on the second phase distribution.The spheroidal structure will be much weakerthan the eutectoid structure. The strength of theeutectoid is a function of cooling rate, faster cooling the plates are narrower and the strength is higher than slow cooling rateswith wider plate spacing.Aluminum alloys – age hardening producesoptimum properties – small particles whichinteract with dislocations very effectively.
Second Phases
Second Phase ParticleDislocation mobile
Dislocation pinned by particle
= Gb/R
= shear stress to keep dislocation movingR = radius of curvature of dislocationsAs R decreased, increases so strengthening the material.R decreases as the particles are closer together, so the distribution is important
New Phases.
Best example would be steel transformation to martensite. Other alloy systems are also
capable of this type of diffusionless transfer such as titanium alloys Ti-6Al-4V, Fe-Ni
alloys. In this case the crystal structure is one that has a very high critical resolved shear
stress such as body centered tetragonal.
Other structures can produce high strength such as “amorphous” or “glassy” metals.
These metal alloys systems are quenched very rapidly, at a rate of several thousand
degrees per second. In this case, the resulting structures are not crystalline and so
have few dislocations and behave elastically to higher yield stresses.
Ni-Ti systems are a good example of these. They are metastable, and cannot be used at
temperature otherwise they gradually revert to their equilibrium crystal structure.
Homework
• It the temperature is increased from 0.5 to 0.8 of Tm (K) what effect does this have on the recrystallization process?
• For aluminum estimate the shear stress required to move a dislocation when R =20nm – use half the elastic modulus as the shear modulus. What is the effect of decreasing the radius of curvature of the dislocation?